Respuesta :
Solution :
Given :
D = the rate of the planned usage of using the work center = 200 parts per hour
T = the average waiting list for the replenishment of the parts plus the average time of production of a container of parts = 0.75 hours
X = the policy variable which is set by the management - the possible inefficiency of the system = 0.11
C = the capacity of the standard container = 6 parts per container
Therefore the total number of the containers are :
[tex]$N=\frac{DT(1+X)}{C}$[/tex]
[tex]$N=\frac{(200 \times 0.75)(1+0.11)}{6}$[/tex]
N = 27.75 which is almost 28 containers.
The rounding up will cause a system to be a looser while rounding down will make the system tighter. Usually the rounding up is preferable.
The number of containers required to support a lean process will be 28 containers.
The formula that can be used to solve the problem will be:
N = DT(1 + X) / C
where,
- D = Planned usage rate = 200
- T = Average waiting time = 0.75 hours
- X = Policy variable = 0.11
- C = Capacity of a standard container = 6
Therefore, the number of containers will be:
N = DT( 1 + X) / C
N = (200 × 0.75)(1 + 0.11) / 6
N = 150(1.11) / 6
N = 166.5/6
N = 27.75 = 28
In conclusion, the correct option is 28 containers.
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